Comments on: Poker Combination Calculation Exercise

By: hgfn klksds

hgfn klksds — Sun, 13 Dec 2009 11:19:27 +0000

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By: David N. Barber

David N. Barber — Wed, 24 Jun 2009 18:02:54 +0000

i have a question. in hold-em, one player has royal flush, the other full house. how many combos are there to get this reasults ??

By: Skipatore

Skipatore — Sat, 05 Apr 2008 09:28:02 +0000

Hairsplitter’s finest hour: Exposed burn cards. If a dealer exposes and burns the AH we have six variations of AAA KK full house board chops. And Dude, let’s not forget, let’s not forget how many (20) cards in the deck, as the exposed burn, can’t complete royal flushes that otherwise hafta chop.(That ain’t legal either.)

By: fabadam

fabadam — Sat, 05 Apr 2008 08:28:55 +0000

@jamleeco:
360 3-flushes is correct. I had “grunched” my answer (mot looking at any answer prior).

By: BadBeatExpert

BadBeatExpert — Fri, 04 Apr 2008 21:36:40 +0000

I used weasel97′s method for calculating the number of straights without flush possibilities (600). Adding in the royal flushes and quads, I also get 656 nut boards, just like weasel97 and Toonces, the driving cat. That means only 0.0252% of boards fulfill this criterion.

I guess the lesson is that we shouldn’t be overly eager to call when we see someone make a big bet on a board with a big made hand (such as a straight) — it’s very possible that it’s a value bet.

Also, let me add that I’m impressed by all of your replies. I hope NOT to see you playing against me in my crappy pokerstars 3.40 SnGs and 25NL games.

By: jamleeco

jamleeco — Fri, 04 Apr 2008 15:02:13 +0000

I just read all the comments. I took the same approcah as Weasel and came up with same answer.

However, since I usually play on weekdays and stayed home from casino today for new showerdoor installation guy, I am supposed to have been cleaning house thus far.

Thanks Ed. =)

By: jamleeco

jamleeco — Fri, 04 Apr 2008 14:53:08 +0000

Fadaham,

I keep coming up with 360 3-flushes. The missing 2 card combinations to broadways =10.

There are 9 ways this can occur without making a 4 flush. The scenario for each of the 4 suits in the 3 flush, and I get 360 ?

By: fabadam

fabadam — Thu, 03 Apr 2008 23:00:42 +0000

What boards apply and how many:
* Royal flushes: there are 4, one in each suit
* Rainbow top straights: AKQJT, no three of each suit. There’s 480 of those (see below).
* xxxxA quads (LOL Kristy Gazes) = 4 x 12 = 48
* AAAAK = 4

Why 480 straights? AKQJT can come up in 4^5=1024 ways. But not counting are:
(*) 4 royals
(*) 60 4-flushes
(*) 480 3-flushes

So there are 4+480+48+4 = 536 nut boards.

The total number of boards is 52x51x50x49x48/5x4x3x2x1 = 2,598,960

So the percentage of nut boards is:
536 / 2,598,960 = 0.0206236 %
That’s about one in every 2 in every 10,000 boards. Which means I should have seen about 50 in my lifetime now, but it seems much more ..

By: Simon

Simon — Thu, 03 Apr 2008 20:22:51 +0000

I think the way to calculate it is much easier. You only have to calculate the broadways straights which is 4^5. This gives us all the A-T combinations including royal flushes): 1024 possibilities. The rest are quads with the nut kicker: kings to deuces with an ace: 12*4. plus Aces with a king: 4 more hands. summa summarum 1076 hands.

By: Ed Miller

Ed Miller — Thu, 03 Apr 2008 19:00:10 +0000

Sorry about the spam filter, guys. It’s got a high false positive rate (and sometimes lets genuine spam through too). If I get rid of it the site will be filled with spam, but I could look into a captcha-style replacement I suppose.

For now, whenever I get a chance, I go through the spam filter and unspam all the good comments.